Feb15-1 - Prob|em 3_59 3.69 The hoilow steel shafi show(6...

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Unformatted text preview: Prob|em 3_59 3.69 The hoilow steel shafi show (6 = 77.2 GPI. Q: = 50 MPa)_r_otales at 240 rpm. Dclmnine (up) the maximum pow that man be Iranmlinad. (b) the corresponding angle of twist of'lhe shaft. c, = 3+3, = {2.5% J = ECCJ— C?) = ,[CsoY'-Ctz5>“] = 1-13HXIO‘ W” = I.Z34 “0“»: T... ‘— SOXID“ Pat _ '1- = 3.3-. -.- QOI‘IO‘HGJH we"): Inn '3? c 30310—3 2056.7 N‘M l: Anjupaw 5,9889! -_ 2'10 ppm ‘-' Ll'r—cv/sec ‘= ‘f' Hz. (5.1 Power berm: wan-«M‘H-eal. P = air-FT: ZECQKQOS'GJ): 51.? via“ w P = 5L"? w) 4 - . _ t. - lose-71(5) _ (L‘! flm§P_¢_____¢F hast (9 - g? - m — c.1073 ml g3 = Li?" Problem 337 337 mmmmfiwmfi-mmmmflfi kW at_aspeeduf2100 rpm. duaminethe minimum mliusroflhe fillet iflm allowable shearingmaf 59 MPais no! to b'ecx'ceeded. loo _ so — 35H: 4: = P v #52103 w —_.___ P :— 45? £05 1 2'"; 117 (35') - For :rmjfier Eu'Je C.='5'ul 1- 15mm = 0.05m a _7r_’t.’r:: _ 1r(50vlo‘3(o.0|s)-" __ K" 2'? ' (2](.‘Z.o‘l.43\ ‘ 1.245 g— = 2 Fm... F123- 3.32 r 0.x? :1 = (0.171803 v 5.1.“ Problem 3 88 333 The W shall shown must transmit 45 kW. Knowingthal the allowable ° simlillgstrminfileshaflistPaandflmthemdiusofthcfillelisr=6m determine the smallust pennissiblc speedofthe shalt. Tl 60mm £_£- .975: d-3o'o'2 a] 30 From F13. 3.32 K': L2G =2 30. mm P c..- -.._- 9L. .- _:_ .- _ .L J_ Ir _ A At..- ...
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